Proof of Cauchy Sequences for a Positive Semidefinite Symmetric Bilinear Form: Cauchy-Schwarz, the Space of Cauchy Sequences, Convergence of Pairings, and Null Sequences

This proof uses: the definitions Vector Space over a Field, Linear Map and Linear Subspace, and claims 1, 2 and 5 of Elementary Identities in a Vector Space; for real numbers, Elementary Order Arithmetic in an Ordered Field, Elementary Arithmetic in an Ordered Field, Properties of the Absolute Value in an Ordered Field, Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field, Existence and Uniqueness of the Nonnegative Square Root and claims 3(c) and 3(e) of Real Powers Through the Exponential, and Elementary Asymptotic Tools: Monotonicity, Null Sequences of Negative Powers, Exponential Domination, Integer Rounding, and Square-Root and Exponential Inequalities; for limits, Limit of a Sequence of Real Numbers, Arithmetic of Limits of Real Sequences, Order Properties of Limits of Real Sequences, Constant Sequences and Index-Shifted Sequences of Real Numbers §constant, claim 1 of Uniqueness of Limits and Boundedness of Convergent Real Sequences, Cauchy Sequence of Real Numbers and Every Cauchy Sequence of Real Numbers Converges; and, for indices, claims 1 to 3 of Properties of the Order on the Natural Numbers (the order of $\mathbb{N}$ is total) with claim 1 of Elementary Properties of the Maximum of Two Elements. Rearrangements of sums, differences and multiples of vectors below are consequences of conditions 1 to 8 of Vector Space over a Field and claims 1, 2 and 5 of Elementary Identities in a Vector Space; in particular $u-v=u+(-1)v$ and $v-u=(-1)(u-v)$. Write $\|\cdot\|$ for $\|\cdot\|_{\beta}$; thus $0\le\|v\|$ and $\|v\|^{2}=\beta(v,v)$ by Existence and Uniqueness of the Nonnegative Square Root.

Step 0 (bilinearity in both arguments). By symmetry, $u\mapsto\beta(u,v)$ is also linear for each $v$. Hence for $u,v\in V$ and $t\in\mathbb{R}$,

Also $\beta(0,v)=\beta(0\cdot0,v)=0$ by claim 3 of Elementary Identities in a Vector Space, so $\|0\|=0$ (the square root of $0$ is $0$, by uniqueness in Existence and Uniqueness of the Nonnegative Square Root).

Step 1 (claim 1, Cauchy-Schwarz). Let $u,v\in V$ and $b=\beta(u,v)$. By claim 1 of Properties of the Absolute Value in an Ordered Field, $|b|$ is $b$ or $-b$, so $|b|^{2}=b^{2}$ by claim 2 of Zero Products and Elementary Identities in a Field. Case $\beta(v,v)=0$: then (1) and positive semidefiniteness give $0\le\beta(u,u)+2tb$ for every real $t$. If $b\ne0$, then $2b\ne0$ (as $0<2$ by claim 8 of Elementary Order Arithmetic in an Ordered Field, and claim 3 of Zero Products and Elementary Identities in a Field), and $t=-(\beta(u,u)+1)/(2b)$ gives $\beta(u,u)+2tb=-1<0$ (claims 6 and 4 of Elementary Order Arithmetic in an Ordered Field), a contradiction; so $b=0$ and $|b|=0=\|u\|\cdot\|v\|$ since $\|v\|=0$. Case $\beta(v,v)\ne0$: then $0<\beta(v,v)$, and $t=-b/\beta(v,v)$ in (1) gives $0\le\beta(u,u)-b^{2}/\beta(v,v)$; multiplying by $\beta(v,v)$ (claim 5 of Elementary Arithmetic in an Ordered Field) and translating (claim 3 there), $|b|^{2}=b^{2}\le\beta(u,u)\beta(v,v)=(\|u\|\|v\|)^{2}$. Since $0\le|b|$ and $0\le\|u\|\|v\|$ (claim 5 of Elementary Arithmetic in an Ordered Field), claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field gives $|b|\le\|u\|\|v\|$.

Step 2 (claim 1, the seminorm). Triangle: by (1) with $t=1$, $b\le|b|$ (claim 3 of Properties of the Absolute Value in an Ordered Field), Step 1 and claim 5 of Elementary Arithmetic in an Ordered Field,

and both $\|u+v\|$ and $\|u\|+\|v\|$ are nonnegative (claim 2 of Elementary Arithmetic in an Ordered Field), so $\|u+v\|\le\|u\|+\|v\|$ by claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field. Homogeneity: $\|cu\|^{2}=\beta(cu,cu)=c^{2}\beta(u,u)=(|c|\,\|u\|)^{2}$, using $|c|^{2}=c^{2}$ as in Step 1; both sides being nonnegative, $\|cu\|=|c|\,\|u\|$ by claim 3 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field. Reverse triangle: $\|u\|=\|(u-v)+v\|\le\|u-v\|+\|v\|$ gives $\|u\|-\|v\|\le\|u-v\|$, and likewise $\|v\|-\|u\|\le\|v-u\|=\|(-1)(u-v)\|=\|u-v\|$; so $-\|u-v\|\le\|u\|-\|v\|\le\|u-v\|$ (claim 4 of Elementary Order Arithmetic in an Ordered Field), which is the reverse triangle inequality by claim 6 of Properties of the Absolute Value in an Ordered Field.

Step 3 (claim 2). A constant sequence $(v)$ satisfies $\|v-v\|=\|0\|=0<\varepsilon$ for all indices. Let $(u_{k}),(v_{k})\in C_{\beta}$, $c\in\mathbb{R}$ and $\varepsilon>0$. Choose $N_{1},N_{2}$ for $\varepsilon/2$ (which is positive, claim 8 of Elementary Order Arithmetic in an Ordered Field) for $(u_{k})$ and $(v_{k})$, and let $N=\max\{N_{1},N_{2}\}$; for $k,l\ge N$ we have $k,l\ge N_{1},N_{2}$ (claim 1 of Elementary Properties of the Maximum of Two Elements, claim 1 of Properties of the Order on the Natural Numbers), and by Step 2 and claims 3 and 8 of Elementary Order Arithmetic in an Ordered Field,

For multiples, $\|cu_{k}-cu_{l}\|=|c|\,\|u_{k}-u_{l}\|$ by Step 2; if $c=0$ this is $0<\varepsilon$, and otherwise $0<|c|$ (claim 1 of Properties of the Absolute Value in an Ordered Field) and choosing $N$ for $\varepsilon/|c|$ (positive by claims 5 and 7 of Elementary Order Arithmetic in an Ordered Field) gives $|c|\,\|u_{k}-u_{l}\|<\varepsilon$ for $k,l\ge N$ by claim 10 there. So $C_{\beta}$ is closed under termwise sums and multiples. Conditions 1, 2 and 5 to 8 of Vector Space over a Field hold in $C_{\beta}$ because they hold in $V$ at every index; the constant sequence $0$ lies in $C_{\beta}$ and satisfies condition 3, and for $(u_{k})\in C_{\beta}$ the sequence $((-1)u_{k})\in C_{\beta}$ satisfies condition 4, since $u_{k}+(-1)u_{k}=0$ for every $k$ (claim 5 of Elementary Identities in a Vector Space). So $C_{\beta}$ is a real vector space with zero vector the constant sequence $0$.

Step 4 (eventual bounds). Let $u=(u_{k})\in C_{\beta}$. Choosing $N_{u}$ for $\varepsilon=1$ and putting $M_{u}=\|u_{N_{u}}\|+1$, Step 2 gives $\|u_{k}\|\le\|u_{k}-u_{N_{u}}\|+\|u_{N_{u}}\|\le M_{u}$ for every $k\ge N_{u}$, and $0<M_{u}$.

Step 5 (claim 3). Let $u,v\in C_{\beta}$, $\varepsilon>0$, and $M=M_{u}+M_{v}$, which is positive. By bilinearity and symmetry, $\beta(u_{k},v_{k})-\beta(u_{l},v_{l})=\beta(u_{k}-u_{l},v_{k})+\beta(u_{l},v_{k}-v_{l})$, so by claim 5 of Properties of the Absolute Value in an Ordered Field, Step 1 and Step 4, for $k,l\ge\max\{N_{u},N_{v}\}$,

Let $\delta=\varepsilon/(2M)>0$ and choose $N'$ with $\|u_{k}-u_{l}\|<\delta$ and $\|v_{k}-v_{l}\|<\delta$ for $k,l\ge N'$ (maximum of two indices as in Step 3). For $k,l\ge\max\{N_{u},N_{v},N'\}$ the right side is $<\delta M+\delta M=\varepsilon$ (claims 3 and 10 of Elementary Order Arithmetic in an Ordered Field). So $(\beta(u_{k},v_{k}))_{k}$ is a Cauchy sequence (Cauchy Sequence of Real Numbers) and converges by Every Cauchy Sequence of Real Numbers Converges; its limit $\widehat\beta(u,v)$ is unique by claim 1 of Uniqueness of Limits and Boundedness of Convergent Real Sequences. Symmetry: $(\beta(u_{k},v_{k}))$ and $(\beta(v_{k},u_{k}))$ are the same sequence. Bilinearity: for $w\in C_{\beta}$ and $c\in\mathbb{R}$, $\beta(u_{k},v_{k}+w_{k})=\beta(u_{k},v_{k})+\beta(u_{k},w_{k})$ and $\beta(u_{k},cv_{k})=c\,\beta(u_{k},v_{k})$, so by claims 1 and 3 of Arithmetic of Limits of Real Sequences and uniqueness of limits, $\widehat\beta(u,v+w)=\widehat\beta(u,v)+\widehat\beta(u,w)$ and $\widehat\beta(u,cv)=c\,\widehat\beta(u,v)$. Positive semidefiniteness: $0\le\beta(u_{k},u_{k})$ for every $k$, the constant sequence $0$ converges to $0$ (Constant Sequences and Index-Shifted Sequences of Real Numbers §constant), so $0\le\widehat\beta(u,u)$ by claim 1 of Order Properties of Limits of Real Sequences. If $u,v$ are constant, $(\beta(u_{k},v_{k}))$ is the constant sequence $\beta(u_{1},v_{1})$, whose limit is $\beta(u_{1},v_{1})$ by Constant Sequences and Index-Shifted Sequences of Real Numbers §constant.

Step 6 (claim 4). Subspace: the constant sequence $0$ has $\|0\|=0$, so it lies in $N_{\beta}$. For $u,z\in N_{\beta}$ and $c\in\mathbb{R}$: $0\le\|u_{k}+z_{k}\|\le\|u_{k}\|+\|z_{k}\|$, where the right side tends to $0$ by claim 1 of Arithmetic of Limits of Real Sequences, so $\|u_{k}+z_{k}\|\to0$ by claim 2 of Order Properties of Limits of Real Sequences; and $\|cu_{k}\|=|c|\,\|u_{k}\|\to0$ by claim 3 of Arithmetic of Limits of Real Sequences. With Step 3 this shows $N_{\beta}$ is a linear subspace of $C_{\beta}$ (Linear Subspace). Characterisation: $\beta(u_{k},u_{k})=\|u_{k}\|^{2}$. If $\|u_{k}\|\to0$ then $\|u_{k}\|^{2}\to0$ by claim 2 of Arithmetic of Limits of Real Sequences, so $\widehat\beta(u,u)=0$. Conversely, if $\beta(u_{k},u_{k})\to0$, then, the terms being nonnegative, claim 3(e) of Real Powers Through the Exponential, and Elementary Asymptotic Tools: Monotonicity, Null Sequences of Negative Powers, Exponential Domination, Integer Rounding, and Square-Root and Exponential Inequalities gives $\|u_{k}\|=\sqrt{\beta(u_{k},u_{k})}\to\sqrt{0}=0$, so $u\in N_{\beta}$. Orthogonality: let $u\in C_{\beta}$, $z\in N_{\beta}$. For $k\ge N_{u}$, Steps 1 and 4 and claim 5 of Elementary Arithmetic in an Ordered Field give $|\beta(u_{k},z_{k})-0|\le\|u_{k}\|\,\|z_{k}\|\le M_{u}\|z_{k}\|$, and $M_{u}\|z_{k}\|\to0$ by claim 3 of Arithmetic of Limits of Real Sequences; by claim 3(c) of Real Powers Through the Exponential, and Elementary Asymptotic Tools: Monotonicity, Null Sequences of Negative Powers, Exponential Domination, Integer Rounding, and Square-Root and Exponential Inequalities, $\beta(u_{k},z_{k})\to0$, so $\widehat\beta(u,z)=0$ by uniqueness of limits.

Step 7 (claim 5). Let $u,v\in C_{\beta}$. If $[u]=[v]$, then $u=u+0\in[u]=[v]$ (as $0\in N_{\beta}$), so $u=v+z$ with $z\in N_{\beta}$ and $u-v=z\in N_{\beta}$. Conversely let $z_{0}=u-v\in N_{\beta}$. For $z\in N_{\beta}$, $u+z=v+(z_{0}+z)$ and $v+z=u+(z-z_{0})$, where $z_{0}+z$ and $z-z_{0}=z+(-1)z_{0}$ lie in $N_{\beta}$ by Step 6; hence $[u]\subseteq[v]$ and $[v]\subseteq[u]$, so $[u]=[v]$. Now let $[u]=[u']$ and $[v]=[v']$, so $u-u',v-v'\in N_{\beta}$. Then $(u+v)-(u'+v')=(u-u')+(v-v')\in N_{\beta}$ and $cu-cu'=c(u-u')\in N_{\beta}$, which gives $[u+v]=[u'+v']$ and $[cu]=[cu']$ by the equivalence just proved. Finally, by bilinearity and symmetry of $\widehat\beta$ (Step 5) and the orthogonality in Step 6,